Calculus (3rd Edition)

Published by W. H. Freeman

Chapter 15 - Differentiation in Several Variables - 15.3 Partial Derivatives - Exercises - Page 781: 51

Answer

$$\frac{\partial W}{\partial E} =-\frac{1}{kT}e^{-E/kT},$$ $$\frac{\partial W}{\partial T} =\frac{E}{kT^2}e^{-E/kT}.$$

Work Step by Step

Since $W=e^{-E/kT}$, then using the chain rule, we have $$\frac{\partial W}{\partial E}=e^{-E/kT} (-1/kT)=-\frac{1}{kT}e^{-E/kT},$$ $$\frac{\partial W}{\partial T}=e^{-E/kT} (E/kT^2)=\frac{E}{kT^2}e^{-E/kT}.$$

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